Optimizing Chemical Process Yield Using Response Surface Methodology

1. Lecture 17 • Today: Start Chapter 9 • Next day: More of Chapter 9

2. Example • Chemical Engineer is interested in maximizing yield of a process • 2 variables influence process yield: • reaction time (x1) and reaction temperature (x2) • Current operating conditions have reaction time at 35 minutes and temperature at 155 oF, which give a yield of about 40% • Best operating conditions may be far from current conditions

3. Response Surface Methods • Often the goal of experimentation is to optimize (say maximize) a system response • When there are only a few quantitative factors, response surfacemethodology can be use for understanding the relationship between the response and input factors • Experimentation strategy is sequential

4. Models • Relationship between input variables and response: • If the expected response is denoted E(y)=f(x1, x2,…, xk), then f(x1, x2,…, xk) is called the response surface

5. Models • First order model to approximate f: • Second order model to approximate f: • Typically a lower order polynomial is used to approximate the local response surface

6. General Idea • If there are many factors to consider, perform a screening design first (e.g., 2k-p fractional factorial design) to screen out unimportant factors • Response surface methodology involves experimentation, modeling, data analysis and optimization • First run a sequence of small experiment designs to fit a first order model to identify the experiment region that is near or contains the optimum • Next an experiment design is performed to estimate the second order model close to the optimum • Designs to estimate the response surface are called response surface designs

7. General Idea • Design that estimates a first order model is called a first order design • Design that estimates a second order model is called a second order design • Analysis is performed using regression

8. Coding the Variables • It is convenient to code the variables (as we have done so far) • For example consider a factor with 3 equally spaced levels • Let zi be the mid-point of the levels • The 3 levels are: Z: zi-c, zi, zi+c • Transformation: • 3 coded levels: X: -1, 0, 1

9. First Order Designs • Will run a series of first order designs until near the optimum • When there is substantial curvature, first order model becomes ineffective for approximating the surface • How do we know if there is curvature?

10. First Order Designs • Must have more than 2 level factors to check for curvature • Solution is to add experiment trials at the center of the experimental region (e.g., x=(0,0,0,0…0) ) • Designs combine factorials and center points

11. Example • Engineer decide that reaction time should be investigated in the area of the operating conditions • time = 35 minutes (z1=30 or 40) • temperature = 155 oF (z2=150 or 160) • Will include center points at the current operating conditions • Coded variables:

12. Example • Design and responses:

13. Curvature Check • Have nftrials at the factorial design points (e.g., -1 and +1 combinations) • Have nctrials at the center point • Motivation for test: • Test:

14. Example • Fit regression line to data, include a quadratic term for curvature check

15. Method of Steepest Ascent (climbing the hill) • Linear effects for first order model estimated by least squares: • Take partial derivative with respect to each variable: • Direction of steepest ascent:

16. Method of Steepest Ascent • Several experiment trials are taken along the line from the center point of the design, in the direction of the steepest ascent until no further increase is observed • The location where the maximum has occurred is the center point of the next first order design • Design should have nctrials at the center point • If curvature is detected, augment the design with additional trials so that the second order model can be estimated

17. Method of Steepest Ascent